这是一个工作代码，我已经在过去：

.data 

result: .space 4
num1: .word -8,num3
num2: .word 1988,0 
num3: .word -9034,num5 
num4: .word -100, num2
num5: .word 1972, num4 

.text

main:

    li   $a0, 128  #numbers of bytes to allocated
    li   $v0, 9    #sys function allocate heap memory
    syscall
    
    la $t1, result
    sw $v0, ($t1)  #store the address of the allocated heap memory in result

    la $t0, num1

while:
    lw $a0, ($t0)

print_value:
  
        lw $a0, ($t0)
    li $v0, 1
    syscall
    
    lw $a0, ($t0)
    jal base10to4
    
    li $a0, '\n'
    li $v0, 11
    syscall

    addi $t0, $t0, 4
    
    lw $t0, ($t0)
        
    beq $t0, 0, end_while 
    
    j while
    
end_while:

end_program:
    li $v0, 10
    syscall

base10to4:

    move $s0,$a0
    la   $s7,result 

    li $a0, '|'
    li $v0, 11
    syscall

    li   $s1, 4
    li   $s2, 0 #index for the result array
    li   $s6, 0 #will keep track if number is negative of positive 
    
divide_loop:
    
    beq  $s0, $zero, end_loop
    
    div  $s0, $s0, $s1
    mfhi $s5
    
    
    move $s3, $s2 
    mul  $s3, $s3, $s1
    lw   $s4, ($s7)
    add  $s3, $s3, $s4
    blt  $s5, $zero, negative
    j    next

negative:
    li   $s6, 1 
    mul  $s5, $s5, -1 # if negative then make positive

next:
    sw   $s5, ($s3)

    addi $s2, $s2, 1 #increment index with 1
    j divide_loop
    
end_loop:   

print_result:
        # $s2 will contain the range of result array
    addi $s2, $s2, -1
    beq $s6, 1, print_minus
        j print_loop
        
print_minus:

    li   $a0, '-'
    li   $v0, 11
    syscall

                          
print_loop: 
    blt $s2, $zero end_print_loop
    
    move $s3, $s2 
    mul  $s3, $s3, 4
    lw   $s4, ($s7)
    add $s3, $s3, $s4
    lw   $s5, ($s3)

    addi $s2, $s2, -1 #increment with 4 for the next address memory location
    
    move $a0, $s5
    li   $v0, 1
    syscall
    
    
    j print_loop
    
end_print_loop:

return_base10to4:

    jr $ra